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Thread: Opthalmic Optics

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    Master OptiBoarder sandeepgoodbole's Avatar
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    Opthalmic Optics

    For 20/20 vision, how much area on Macula or Retina ,in Square Millimeters or Micrometers is enough ? When we add an error of minus 1 diopter, whats the diameter of the diffused focus beam projected over the Retina ?

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    Master OptiBoarder OptiBoard Silver Supporter Barry Santini's Avatar
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    Sorry, I don't know that or how to calculate it.

    I have read, though, that if the "average" eye's axial length is increased approx 0.3mm, that this is equal to 1.00 diopter of myopic defocus. If this is correct, then the "smallest discernable difference" (as the ANSI committee refers) of 0.25D is equivalent to a Optical Path Difference
    (OPD) of 0.075mm.

    Darryl?

    Barry

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    Quote Originally Posted by sandeepgoodbole View Post
    For 20/20 vision, how much area on Macula or Retina ,in Square Millimeters or Micrometers is enough ? When we add an error of minus 1 diopter, whats the diameter of the diffused focus beam projected over the Retina ?
    They don't correlate to each other the way in which you would think. If you were to give a particular size of an object and then asked what size the area on the retina would need to be to see the object then the question may be answered. You figure 20/20 correlates to 5 minutes of arc at 20 feet, so the letter E on the chart has each contrasting space covering about a minute of arc making it discenranble. My guess is that for an E you could probably go down to a point where each minute of arc projects it's image onto one cone in the macula, if that were the case then the width of 5 retinal cones could be the minimum amount of space to see 20/20 although I believe I have read somewhere that stimulation of the cones would occur when light strikes multiple cells, now if that were the case then the minimum amount of retinal area to see an E would need to be the width of the minimum amount of retinal cells that are stimulated by 1 minute of arc * 5.

    I hope that made sense.

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    Carl Zeiss Vision OptiBoard Gold Supporter Darryl Meister's Avatar
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    In a typical eye, the cones in the fovea are roughly 1.5 microns (0.0015 mm) in diameter. According to Hartridge's criterion, two object points can only be resolved as separate if there is at least one lesser stimulated cone between two stimulated cones.

    This would require a separation between the centers of the outer two stimulated cones of at least 1.5 * 2 = 3 microns. Now, the visual angle subtended by these two image points on the retina will depend upon the length of the globe.

    When measuring the visual angle measured at the nodal points of the "schematic" eye, the distance from the fovea to the second nodal point is roughly 16.5 mm. Consequently, the visual angle A subtended by a 3 micron (0.003 mm) separation between the centers of two retinal cones is given by:

    A = 0.003 mm / 16.5 mm = 1.8182 radians (0.0104 degrees or 37.5'' of arc)

    This is equivalent to 20/12.5 visual acuity. In reality, ocular aberrations and other factors limit the maximum potential vision quality of the eye, resulting in visual acuities that are slightly worse than 20/12.5.
    Darryl J. Meister, ABOM

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